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If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions? Please express your answers with 4 decimal place. _________________________ m

Respuesta :

Answer:

[tex]L_f = 0.0196[/tex]

Explanation:

given,

mass of the car,m = 0.2 Kg

height to width of ramp, y/x = 12/75

initial displacement, L_i = 2.25 m

change in momentum, Δp = 0.58 kg.m/s

distance of change in direction of the ramp, L_f = ?

Using equation

[tex]\Delta P = m \sqrt{\dfrac{2gy}{x}} (\sqrt{L_i}+\sqrt{L_f})[/tex]

inserting all the values

[tex]0.58 = 0.2\times \sqrt{\dfrac{2\times 9.8\times 12}{75}} (\sqrt{2.25}+\sqrt{L_f})[/tex]

[tex]1.5 + \sqrt{L_f} = 1.64[/tex]

[tex]\sqrt{L_f}=0.14[/tex]

[tex]L_f = 0.0196[/tex]

The car will cost back up to the distance of 0.0196 m

okpalawalter8

We have that the  distance coast back up the ramp   is mathematically given as

 [tex]\sqrt{lf}=-0.3406[/tex]

From the question we are told

  • If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m,
  • and the change in momentum is 0.58 kg*m/s,
  • how far will it coast back up the ramp before changing directions?

Distance

Generally the equation for the momentum  is mathematically given as

[tex]dm=m\sqrt{\frac{2gy}{x}}*(\sqrt{l1}+\sqrt{lf})\\\\Therefore\\\\0.48=0.2\sqrt{\frac{29.8*12}{85}}*(\sqrt{1.45}+\sqrt{lf})[/tex]

Hence

  • [tex]\sqrt{lf}=-0.3406[/tex]

For more information on momentum  visit

https://brainly.com/question/22568180

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